Mathematics - 2014-02-04 (page 2 of 3)

N3buchadnezzar

12:00

So then we are left with $\text{Re}\left( \int_0^{2\pi} e^{x} \mathrm{d}x \right)$

Sush

@robjohn, please help me too!

usukidoll

12:22

going sleep night

skullpatrol

later

Studentmath

Can I say that the ordinal of seg(b) where b belongs to B, has an ordinal equal or lesser to that of B? (where both are ordered in the same well-order)

Karl Kronenfeld

Not sure what seg(b) is, or B for that matter

Studentmath

Say that <A,<> is an ordered set, the set seg(a) [where a belongs to A] is the set {x belongs to A: x<a}

Which means the set that includes all the 'items' in A that come before the item a

Karl Kronenfeld

ok

yes, you can say that

Studentmath

12:35

Now, I want to say that the ordinal of <seg(a),<> is lesser or equal to the ordinal of <A,<>.. I need it as a lema for some proof, not sure how to go about proving that though

I see, thought so - any hint on how to go about proving it though?

Actually, I will go back to the definition of ordinal and go about figuring it out that way. Thanks Karl!

Karl Kronenfeld

Yeah, it is immediate from the definition.

(Or, at least, the definition I have in mind)

Studentmath

It's not the definition I have in my text-book, that's for sure :P

Karl Kronenfeld

You say two well-ordered sets are equivalent if and only if there is an order-preserving bijection between the two.

Studentmath

It is the same definition.

Karl Kronenfeld

Then, each equivalence contains a unique von-Neumann ordinal.

Studentmath

12:41

I am not yet into von-Neuman ordinals, unfortunately.

Karl Kronenfeld

You can also order the equivalence classes themselves, so no big deal.

robjohn

@N3buchadnezzar it is now a contour integral...

Karl Kronenfeld

The ordering is given by $\alpha<\beta$ iff there exists $A\in\alpha$ and $B\in\beta$ such that $A$ is an initial segment of $B$. (Where $\alpha,\beta$ are equivalence classes, as above.)

Studentmath

So I get that <seg(a),<> has a unique ordinal, and so does <A,<>, question is how do I bring it about that the ordinal of the first is lesser or equal to the ordinal of the latter

robjohn

@N3buchadnezzar $\mathrm{Re}\left(\oint ze^z\frac{\mathrm{d}z}{iz}\right)$ where the contour is the unit circle.

@Sush I was working on your answer. It is not easy to figure out how best to present some things.

Studentmath

12:46

Probably coming of as stupid here Karl, but wouldn't it be a<=b (= in the case that A is the whole of B)

Oh wait, didn't read the edit..

Karl Kronenfeld

Yeah, it should be.

Too late to fix.

Studentmath

Alright, thanks - that is what I need to prove though, isn't it?

robjohn

@N3buchadnezzar and there is no singularity of $\frac1ie^z$ in the unit circle.

Studentmath

Or well, it's what I want to prove..

Karl Kronenfeld

It serves as a definition. I am not sure how the ordering relation on ordinals is defined in your book, so I just gave the one I was using.

Studentmath

12:51

Between every equivalence class and the set that has that equivalence class there is a bijective function, correct?

Karl Kronenfeld

uh, I can read that in two ways

Studentmath

Oh, oh, nevermind. Got it, finally.

It's not a definition (what you wrote) in my book but can very easily reach to it from three seperate definitions - thanks Karl!

Karl Kronenfeld

No problem

Studentmath

I think I am losing it, it starts to feel fun..

robjohn

@RamanaVenkata The dirac measure is singular with respect to Lebesgue measure. If the set contains $0$ it has dirac measure $1$, if not, it has measure $0$.

@RamanaVenkata can you compute the function $F$ for that measure?

Alexander Gruber

13:09

@BalarkaSen no, i haven't

Studentmath

13:34

Got to another lema, yet not sure if this one is even right.. <A,<> Is an ordered set. <A,<> is well-ordered iff for every a in A, the segment defined by a is well-ordered by <

Balarka Sen

By the way, Alexander, I have seen your answer here. I don't think this is what OP wants (which is not clear, as I have said before).

Studentmath

Any input will be appreciated - I fear it is wrong, since I can imagine a situation where it isn't so clear that it is right, yet it should be right..

Balarka Sen

@AlexanderGruber I am trying to give a functional extension answer to the question. Do you think it worths it?

Studentmath

seg(a)={x belongs to A|x<a} which means all the items before a

Karl Kronenfeld

Yeah, I thought for some reason seg(a) required A to be well-ordered. meh.

Alexander Gruber

13:36

@BalarkaSen sure, by all means

Studentmath

That would've made this one easier..

Karl Kronenfeld

@Studentmath Hm, I am not sure about this one.

Studentmath

Same. It should be true though..

Though, say I take densely ordered set A..

Karl Kronenfeld

I think we could let $A=\{a,b\}$ and put the following trivial ordering relation on it $\{(a,a),(b,b)\}$.

Oh, yeah, you probably want, at a minimum, that $A$ is linearly ordered.

Actually, the ordering relation being linear is good enough, as far as I can tell.

Studentmath

Yeah, if I can get from there to the fact that the ordering relation is linear, then <A,<> is well-ordered

Karl Kronenfeld

13:40

Well, the $A$ I gave is densely ordered.

It's not necessarily linear.

Balarka Sen

@AlexanderGruber I am preparing a draft. The question is so vague that I can't say surely if I am writing what he wants. Review it for me when I finish, will you?

Studentmath

What does that trivial relation mean? (I sound extremely stupid, I know...)

Karl Kronenfeld

I meant a<=a, b<=b and nothing else.

Studentmath

I see, alright

Karl Kronenfeld

You can take two disjoint copies of $\mathbb Q$ for a less contrived example.

Studentmath

13:44

Yes, I see.. it seems to me from that, that the lema isn't true - doesn't it?

Karl Kronenfeld

As stated, the lemma is false.

Studentmath

Odd, but thanks!

Karl Kronenfeld

The given $A$ may have enough properties to make that kind of thing work though.

Studentmath

Yeah another easy example would be a densely ordered set A with a least element, a subset of it without the least element wouldn't necesserily have a least element too.. so it's false. I will look that through though, maybe I am missing other properties that could help me with it, yes. Once again, many thanks!

Karl Kronenfeld

You're welcome

Studentmath

13:50

No, it's true! figured out why too.

Balarka Sen

@Alexander Would you do me a favour?

Keep the question open.

Studentmath

Each segment is well ordered by <. so for every a that belongs to A we take, the segment is well-ordered, so for every given a, all the items before him are well ordered by <, so the entire set is well ordered by <.

Balarka Sen

(I though mods might close it due to vagueness)

Studentmath

(A rough explanation but I hope you get the point)

@KarlKronenfeld

Alexander Gruber

@BalarkaSen i mean, i don't control who accepts what

Balarka Sen

13:57

@AlexanderGruber At least you are a mod. Keep your vote out of that ;)

Alexander Gruber

why would I unilaterally close a question that I answered? Lol

Balarka Sen

LOL

Studentmath

14:18

Plus we speak of ordered sets a.k.a. strictly ordered sets, which is a very important note too

user20997

14:49

what is a topological product of S^1 and R? Does it look like an infinite cylinder?

Mike

yop

robjohn

15:22

@AlexanderGruber yeah, leave that to me :-)

@user20997 That's what it looks like as far as I can see...

N3buchadnezzar

Heya

Balarka Sen

@robjohn NOOO!

I am still preparing the draft!

robjohn

@BalarkaSen If you're worried about people closing the question, post a preliminary answer. You can always edit an answer, even if the question is closed.

The preliminary answer should be a good answer, though

N3buchadnezzar

@robjohn Heya! I understood the method you showed me for integrating the function, but can you check if my following argument holds? Otherwise I will probably ask it as a question on the site

robjohn

@N3buchadnezzar Sure.

N3buchadnezzar

15:27

$u(x,y) = \text{Re}(z e^z) = x e^x \cos y - y e^x \sin y$

From before we know that there exists an analytic function $f = u + i v$

Now we integrate $f$ around a closed path, the unit circle.

Balarka Sen

@robjohn I doubt a preliminary answer would be quite satisfactory. Besides, I am already finished.

N3buchadnezzar

Since $f$ is analytic this integral should be zero

robjohn

@BalarkaSen cool

@N3buchadnezzar yep.

N3buchadnezzar

$$ \oint_C f\,\mathrm{d}s = \oint_C u\,\mathrm{d}s + i \oint_C v\,\mathrm{d}s $$

robjohn

@N3buchadnezzar except, even if the integrand is analytic, the differential is $\frac{\mathrm{d}z}{iz}$

@N3buchadnezzar This picks up the value at $0$ times $2\pi$

N3buchadnezzar

15:31

Since the integral is zero we have that
$$
\oint_C u \,\mathrm{d}s = \frac{1}{i} \oint_C v\,\mathrm{d}s
$$
The lefthandside is real, and the righthandside is imaginary. Hence in order for the two sides to be equal, both integrals has to be zero.

Can one use this argument?

Balarka Sen

@robjohn Can an answer of a closed question have upvotes?

robjohn

@N3buchadnezzar Another way of doing this is to use the mean value property of harmonic functions.

@BalarkaSen and downvotes.

N3buchadnezzar

Yeah, our proffesor told is this was the way to do it

robjohn

@N3buchadnezzar They both amount to exactly the same thing in the case of a circular contour.

Balarka Sen

@robjohn I have finished the essential parts of the question and posting it, leaving a remark that the other part is 'on preparation'. Is that good?

@robjohn You have a cruel sense of humor.

N3buchadnezzar

15:33

But is the line integral method correct as well ? Okay

robjohn

@BalarkaSen sounds fine

N3buchadnezzar

$$
u(z_0) = \frac{1}{2\pi} \int_0^{2\pi} u(z_0 + r e^{i \theta})\,\mathrm{d}\theta
$$

robjohn

@N3buchadnezzar The mean value property is the line integral

@N3buchadnezzar that's it

N3buchadnezzar

Which is the middle value theorem written up for explicit the interval 0 to 2pi

But my problem is that we have $e^{i\theta}$ ?

$h(\theta) = u(\cos x, \sin x)$ which is real

Balarka Sen

@robjohn Fine. Done.

0

A: What is the difference between a non-solvable algebraic number and a transcendental number?

Abel-Ruffini theorem, which I am sure you are familiar with, simply says that the roots of a general polynomial of degree $n \geq 5$ is not solvable in radicals. I will modify this argument a bit to get my answer started. Abel-Ruffini theorem : The roots of a general polynomial of degree $n \geq...

robjohn

15:36

@N3buchadnezzar You simply integrate the harmonic function around the unit circle and get the value at the center.

N3buchadnezzar

Gimme a second to write

robjohn

@BalarkaSen Is $x^5+x+1$ non-solvable? I think that gives one example

Balarka Sen

@robjohn $x^5 + x + 1 = (x^2 + x + 1)(x^3 - x^2 + 1)$ is solvable. And what example?

N3buchadnezzar

$$ \begin{align*}
\int_0^{2\pi} h(\theta)\,\mathrm{d}\theta
& =
\int_0^{2\pi} u(\cos x,\sin x)\,\mathrm{d}x \\
& =
2\pi \text{Re}\left[ \frac{1}{2\pi} \int_0^{2\pi} u(0 + e^{i\theta})\,\mathrm{d}\theta\right]
\end{align*}$$

Balarka Sen

Nevermind.

N3buchadnezzar

15:39

@robjohn Something like this ?

robjohn

@BalarkaSen ah, it's not even irreducible. However, if we have a fifth degree polynomial with a non-solvable Galois group, it can't be solved by radicals, right?

Balarka Sen

@robjohn Yes. So?

robjohn

@N3buchadnezzar That looks right.

@N3buchadnezzar and what is $u(0,0)$?

N3buchadnezzar

zero ofcourse..

robjohn

@BalarkaSen Then that root is a non-solvable algebraic, but not a transcendental.

Balarka Sen

15:41

@robjohn Yes, but I believe the question asks something different.

N3buchadnezzar

I just do not see why we can write $u$ on that form

Balarka Sen

More like a discrimination than direct comparison.

robjohn

@BalarkaSen Oh, I was extrapolating from what I can see here.

N3buchadnezzar

I tried expanding and computing the real part, but they are not alike

robjohn

@N3buchadnezzar Hang on...

N3buchadnezzar

15:42

$u(0 + e^{i}\theta))$ does not even make sense to me, since $u$ should deppend on two variables not one

robjohn

@N3buchadnezzar $e^{i\theta}=(\cos(\theta),\sin(\theta))$

You do know the relation between $\mathbb{C}$ and $\mathbb{R}^2$

N3buchadnezzar

Yeah, but more familiar with $e^{i\theta} = \cos \theta + i \sin \theta$

@robjohn Ah, they are isomorphic

robjohn

@N3buchadnezzar That is how the Cauchy-Riemann equations are imposed

N3buchadnezzar

Thanks, that was the last piece I was missing. Sorry for asking so many questions for each problem. I really try to understand it before moving on.

robjohn

Since they are defined on $\mathbb{R}^2$

Julien

15:54

Hi, someone to confirm :u^2+v^2 is a prime number which is congruent to 1 (mod 4) for this math.stackexchange.com/questions/659023/…

please, thanks

N3buchadnezzar

16:19

=)

agent154

Is anybody here familiar with Java and the Monte Carlo method of estimating $\pi$?

agent154

16:46

I hate stack overflow... They seem like a bunch of stuck up assholes who do nothing but downvote your questions.

There seems to be no point in asking any questions there, as they always effectively tell me to stop asking bad questions and go research it myself... But what if I did research it myself and still can't understand it? Who am I supposed to ask if I can't ask anybody for help?

Studentmath

Last question for me, today: Any good ideas on how to prove that the sum of all Aleph ns where n is natural number is aleph w? I am asked to compute, I know it's the answer, I am just unsure about how to prove it - perhaps by stating the set of such alephs has the ordinal of w, but.. I am unsure how that leads me there, can't find a good proven statement to use or think of one to prove. Any input will be loved!

@agent154 I assume they think such questions belong to stackexchange, and not to professional post-graduate more discussion orienated math forum.. but I may be wrong, never posted that (out of my league)

agent154

@Studentmath I said stack overflow, not math overflow

Studentmath

Oh.

My bad.

agent154

stack overflow is part of stack exchange. It's the site dedicated to programming

Studentmath

Hah, good to know not to ask questions there when I get stack in my programming courses :P

Balarka Sen

16:57

Finally, done.

0

A: What is the difference between a non-solvable algebraic number and a transcendental number?

Abel-Ruffini theorem, which I am sure you are familiar with, simply says that the roots of a general polynomial of degree $n \geq 5$ is not solvable in radicals. I will modify this argument a bit to get my answer started. Abel-Ruffini theorem : The roots of a general polynomial of degree $n \geq...

@AlexanderGruber

@Mike You are betting against something that has been searched upto $10^{165}$?

6

Q: Do we know a number $n\gt 5$ with no twin prime $n\lt q\lt 2n$?

This is essentially a Bertrand's postulate version for twin primes. I am interested in both an explicit example and large lower bounds for it because of this answer of mine. In the comments below the answer, it is shown that there is no such $n$ below $8\times 10^{15}$. An efficient algorithm wo...

Mike

@Balarka I take it back since it would contradict the k-tuple conjecture (obviously, in retrospect), which I'd rather believe than not. But that's got nothing to do with how much we've checked it.

Balarka Sen

@Mike Okay.

I have no real intuition on k-tuple, though, so no betting on that.

small ks are obviously true, not sure what happens at large.

All I can say is that I don't believe in Hypothesis H.

:13582604 Can you explain me the question?

Pedro Tamaroff

17:16

@BalarkaSen I just think it diverts from the question.

Balarka Sen

@PedroTamaroff Yeah, but what is the question?

Pedro Tamaroff

@BalarkaSen "What is the difference between a non-solvable algebraic number and a transcendental number?"

Balarka Sen

@Mike But then I don't believe in H-L convexity conjecture, so I am in rather leagues with k-tuple.

@PedroTamaroff My believe is that this is not what he wants to be answered, which can be easily done as "By definition, of course!"

See his Q before the edit.

Pedro Tamaroff

Don't you need to change something here $$\left[\frac{\vartheta_{10}(0|\tau)}{\vartheta_{00}(0|\tau)}\right]^4 +\left[\frac{\vartheta_{10}(0|\tau)}{\vartheta_{00}(0|\tau)}\right]^4=1$$

?

You wrote $_{10}$ on both terms.

Balarka Sen

@PedroTamaroff Done.

@Pedro Well, if you think the answer is irrelevant, I can delete that. The question is very vague and I knew from the beginning I don't understand what OP wants.

Pedro Tamaroff

17:22

@BalarkaSen I don't know if it is irrelevant, but sometimes it is easy to get carried away.

Balarka Sen

@PedroTamaroff Like where?

Where actually do you think has a tendency of getting carried away?

Mats Granvik

http://mathoverflow.net/a/102186/25104
There is a new plot of the zeta zero counting staircase, remains to do it by integration and not the with the accumulate Mathematica command.

scale = 400;
Print["Counting to 60"]
Monitor[g1 =
ListLinePlot[
Accumulate[(1 -
Table[Re[
Zeta[1/2 - I*k]*
Total[Table[
Total[MoebiusMu[Divisors[n]]/
Divisors[n]^(1/2 - I*k - 1)]/n/Log[scale]/1.35, {n,
1, scale}]]], {k, 0 + 1/1000, 60, N[1/6]}])^12],
DataRange -> {0, 60}, PlotRange -> {-0.15, 15}];, Floor[k]]
Show[g1]

Balarka Sen

@Pedro For example, is Myersen's answer relevant?

Pedro Tamaroff

17:43

@BalarkaSen Gerry?

Chris's sis

@N3buchadnezzar I have something for you I just created. Does it possibly have a nice closed form?

$$1-\zeta(2)+2-\zeta(2)-\zeta(3)+3-\zeta(2)-\zeta(3)-\zeta(4)+\cdots$$

Balarka Sen

@Chris'ssis Likely.

@PedroTamaroff Yes.

Chris's sis

@BalarkaSen Do you have in mind a possible answer?

Balarka Sen

@Chris'ssis Give it to me in series form.

Pedro Tamaroff

@Chris'ssis I don't see a pattern.

@BalarkaSen Well, he seems to be more simplistic: "If the group is solvable, then the roots of the polynomial can be expressed in radicals; if not, not."

Mike

17:46

@PedroTamaroff $$\sum_{n=1}^\infty \left(n-\sum_{k=1}^n \zeta(k+1)\right)$$

Balarka Sen

@PedroTamaroff That's something comment-sized, I think.

Mike

err

where'd that $1^n$ come from

Balarka Sen

@Mike What?

Mike

there we go

Balarka Sen

Pretty good.

Pedro Tamaroff

17:48

Oh. I see.

Balarka Sen

@Chris'ssis interchange summation.

Pedro Tamaroff

@BalarkaSen Justified by what?

Mike

um

the inner sum depends on $n$

Balarka Sen

@PedroTamaroff That can be though of later.

Mike

no, it really can't

Pedro Tamaroff

17:49

@BalarkaSen That's a physicist's talk!

Balarka Sen

@Mike So? What stops?

Pedro Tamaroff

@BalarkaSen OK, just continue with your idea.

Balarka Sen

$$\sum_{n=1}^{\infty}\sum_{k=1}^\infty f(n, k) = \sum_{k=1}^{\infty}\sum_{n=k}^\infty f(n, k)$$

Under some conditions, of course.

Pedro Tamaroff

Sure.

So, you have $$\sum\limits_{n = 1}^\infty {\sum\limits_{k = 1}^n {\left( {1 - \zeta (k + 1)} \right)} } $$

Balarka Sen

I wouldn't work with that if I were you.

Pedro Tamaroff

17:54

Well, the inner terms are all positive.

So you can change the order of summation.

Balarka Sen

Oh, I see.

Sure, then.

Studentmath

Odd. I keep getting something other than I should. Is it possible the number of all countable ordinals who're not follow ups (as in, they can't be represented in a+1) is Aleph_null and not Aleph_one? It's the group {w,w+w,w+w+w,....w*n....}, so it has the ordinal of w, which means it's alike to N ordered by <... what am I doing wrong here?

Balarka Sen

@Chris'ssis Does this help $$\sum_{n=1}^{\infty}\sum_{k=1}^n f(n, k) = \sum_{k=1}^{\infty}\sum_{n=k}^\infty f(n, k)$$

Mike

@Pedro all negative, you mean

2

Pedro Tamaroff

@Mike Yes, nevertheless, it's OK.

Take the effing minus out.

Mike

17:56

right

Pedro Tamaroff

=D

Mike

why am i getting re-involved in this

Pedro Tamaroff

At any rate, I have two group theory problemsto solve.

Studentmath

What's the difference between set and group theory, if I may ask and not sound overly stupid?

Balarka Sen

@Studentmath Groups are sets endowed with binary ops.

Studentmath

17:58

I see

Pedro Tamaroff

@BalarkaSen Well, special binary operations.

Like Ralph.

Balarka Sen

@Pedro math.stackexchange.com/questions/657168/…

How's that? Irrelevant, eh? =D

Pedro Tamaroff

You get used to praise here.

People are too easily impressed.

Balarka Sen

Yeah =D

And that's bad, right?

Pedro Tamaroff

Don't know if it's left or right, just true.

Balarka Sen

18:03

@PedroTamaroff Not everywhere here! Just in Galois theory and transcendental number theory =D

Ian Mateus

@Mike I think the disproof of Bertrand's postulate for twin primes doesn't make the $k$-tuple conjecture false. Hardy just suggested an asymptotic, it says nothing about sporadic irregularities (though it would be a spetacularly strong one)

Balarka Sen

Nimza

Hello

Pedro Tamaroff

@Nimza Hello.

Nimza

@PedroTamaroff hi :)

robjohn

18:18

@BalarkaSen at what rep does an SE user become emeritus?

Balarka Sen

@robjohn 100k. Bill has already become one.

robjohn

There are a number of 100K+ users. There is even two 200K+ users

Balarka Sen

Hm. Then 500k?

Zibadawa

Kind of a catch 22 there. Hit 100k rep, and you are automatically declared retired and inactive. Intentionally stop acquiring rep before you hit 100k, and you are retired and inactive by definition.

Jasper Loy

For me, I delete my accounts periodically, lol.

Balarka Sen

18:21

@JasperLoy How many have you deleted so far?

Jasper Loy

@BalarkaSen I lost count.

Pedro Tamaroff

Jasper has a phoenix complex.

2

Zibadawa

Yes, I suppose the academic system doesn't have the analogue of reincarnation.

Unless you believe in reincarnation, I guess.

Chris's sis

@BalarkaSen what is the final answer you got?

Balarka Sen

@Chris'ssis I haven't tried yet.

Chris's sis

18:25

@BalarkaSen ah, OK.

Pedro Tamaroff

@Mike Pete hasn't answered my mail yet. =(

Balarka Sen

Can we switch like this $$\sum_{k=1}^n \sum_{i=1}^\infty \frac{1}{i^{k+1}} = \sum_{i=1}^\infty \sum_{k=1}^n \frac{1}{i^{k+1}}$$?

Ignore that.

@robjohn?

@Pedro?

Mike

@Ian There would be no tuple of the form $(p,p+2,p+6,p+8)$

Pedro Tamaroff

@BalarkaSen What?

Mike

@Pedro Maybe he never will . . .

Balarka Sen

18:31

2 mins ago, by Balarka Sen

Can we switch like this $$\sum_{k=1}^n \sum_{i=1}^\infty \frac{1}{i^{k+1}} = \sum_{i=1}^\infty \sum_{k=1}^n \frac{1}{i^{k+1}}$$?

Pedro Tamaroff

@BalarkaSen You're a big boy, what do you think?

Balarka Sen

I have no idea. My mind has gone blank.

robjohn

@BalarkaSen looks okay

Balarka Sen

@robjohn Thanksies.

Ian Mateus

@Mike why? Note that I'm assuming a single counterexample suffices to destroy the postulate

Pedro Tamaroff

18:32

@Mike Damn you! =D

Mike

For $p$ large enough, if Bertrand's postulate is false, then $(p,2p)$ has no twin primes

But $p+6$ is decidedly a twin prime if it's part of such a $k$-tuple

Ian Mateus

@Mike this is stronger than the negation of the postulate

Mike

No, it's not

Daniel Fischer

@Studentmath $\omega\cdot\omega,\, \omega\cdot\omega+\omega,\,\dotsc$

Mike

Goddamnit

Pedro Tamaroff

18:36

@Mike I am trying to prove that an abelian group of composite order is not simple. My idea is to show that if $|G|=mn$ then the subgroups $\{x:x^m=1\}$ and $\{x:x^n=1\}$ are proper and nontrivial. In fact, it suffices to do it with one $m$, and we may assume $m$ is prime. Ah, then it is a consequence of Cauchy's theorem.

@DanielFischer HAI!

Mike

Wait no that wasn't worth a goddamnti

Daniel Fischer

@PedroTamaroff Shark? Where?

Mike

@Ian If BP were false for twin primes, then for large enough twin prime $P$, there are no twin primes between $(p,2p)$ (with $p>P$)

Balarka Sen

@Chris'ssis I'll let you simplify $$S = \sum_{i=1}^{\infty} \frac{1}{i^2}\frac{1-1/i^n}{1-1/i}$$

Mike

Now the $k$-tuple conjecture implies there are an infinity of $(p,p+2,p+6,p+8)$

Balarka Sen

18:38

Then you can try computing $\sum_{n \geq 1} (n - S)$

Which will result your summation.

This was quite easy.

Mike

In particular, there's a $(p,p+2,p+6,p+8)$ with $p>P$, which would contradict the falseness of twin-BP

Pedro Tamaroff

@DanielFischer WAT?

Ian Mateus

@Mike ok, now I understand your version of the postulate. I was confused because I added the restriction $p\gt 5$, so my version could have had the scenario "a single counterexample, while still holding for sufficiently large $p$". This doesn't happen in your version, I get it now

Daniel Fischer

@PedroTamaroff "Hai" is German for shark.

Mike

@Ian I don't understand what you mean there. I would implicitly add that too, since there are obviously no twins between $3$ and $6$

Pedro Tamaroff

18:42

@DanielFischer Oh. Didn't know that.

Does anyone know how can I add my yahoo account to my gmail account?

Ian Mateus

@Mike your version: "BP holds for sufficiently large $p$". My version, a stronger one: "BP holds for any $p\gt 5$"

Mike

no

i'm using yours

Daniel Fischer

$\text{yahoo account} + \text{gmail account}$, @PedroTamaroff?

Mike

my "twin BP": "for a prime $p>3$, if $p+2$ is prime, then there exists some prime $p' \in (p,2p)$ with $p'+2$ also prime"

oh

Pedro Tamaroff

@DanielFischer ¬¬ I was told I can make my google account "pick up" my yahoo correspondence too, i.e. add my account to google's gmail.

Mike

18:44

nope, you're right

i'm using the one you said is mine

i redact and return to my original "goddamnit" and storm off

Balarka Sen

@Mike Yep.

@Mike You stormed and returned.

Pedro Tamaroff

$Q_8$ is solvable, right @DanielFischer?

Balarka Sen

Surely.

Chris's sis

@BalarkaSen I don't know what is the easy part. I think something is wrong in my question.

Daniel Fischer

@PedroTamaroff All groups of order $< 60$ are solvable (nice exercise, that, by the way).

Pedro Tamaroff

18:50

@DanielFischer Heh, OK. I have to find all 7 composition series for $D_8$; I found the three composition series for $Q_8$.

Balarka Sen

@Chris'ssis I am currently busy thinking another problem, I will get back after I am done. Meanwhile, @robjohn might jump in?

Daniel Fischer

@PedroTamaroff A composition series, was that with abelian factors, or with simple factors?

Pedro Tamaroff

@DanielFischer Simple.

robjohn

@BalarkaSen That is $\zeta(2)+\zeta(3)+\dots+\zeta(n+1)$

Daniel Fischer

@PedroTamaroff And does your $D_8$ have $8$ elements, or $16$?

Pedro Tamaroff

18:53

@DanielFischer $8$ elements, acts on $4$ vertices.

Balarka Sen

@robjohn I was thinking of partial sums when I wrote that.

Daniel Fischer

@PedroTamaroff Then you don't really have many possibilities to try, should be doable.

Balarka Sen

And taking $\lim_{N \to \infty}$

Pedro Tamaroff

@DanielFischer I have the lattice of subgroups, and I know what the normal subgroups of $D_8$ are, so that's a start.

Daniel Fischer

It doesn't have too many subgroups of index $2$, for a start.

Pedro Tamaroff

18:55

@DanielFischer Three.

And all other subgroups are contained in those.

Daniel Fischer

One of the three has a single subgroup of index $2$, the other two have three each, makes $7$.

robjohn

@BalarkaSen If you start your sum at $i=2$ then the limit is $1$

Pedro Tamaroff

@DanielFischer Right.

Chris's sis

@BalarkaSen this is the correct version $$1+\sum_{n=2}^{\infty}\left(n-\sum_{k=2}^{n}\zeta{(k)}\right)$$

@BalarkaSen I create lots of questions, and there are also mistakes, I'm not a robot. :D

robjohn

@Chris'ssis don't you want $n-1-\sum\zeta(k)$?

Chris's sis

19:02

@robjohn yeah, why not ? With a "+" you mean?

robjohn

@Chris'ssis The thing I was thinking of was $$\sum_{k=2}^\infty(\zeta(k)-1)=1$$

Chris's sis

@robjohn Indeed.

Studentmath

@DanielFischer Yes, thanks, I got to it eventually.. a bit of twisting and few lemas (as I have to use specific proofs in the text-book)

Jasper Loy

@robjohn Other than amazon.com is there any other big seller in the US that has many math books? I know about barnesandnoble.com.

robjohn

@Chris'ssis The sum is of $n-1$ $\zeta$s and I think we need to cancel a $1$ per each

@JasperLoy Those are the two places I would look.

Chris's sis

19:05

@robjohn True, I misunderstood things.

Jasper Loy

@robjohn OK, do you have a preference for one over the other?

Chris's sis

Here is the beauty $$1+\sum_{n=2}^{\infty}\left(n-\sum_{k=2}^{n}\zeta{(k)}\right)=\frac{\pi^2}{6}$$

:D

robjohn

@JasperLoy I don't really. I would just look to see who has your book and if both do, who has it for less.

@Chris'ssis never mind

@Chris'ssis why not just sum $n$ from $1$ instead and forget the extra $1$?

Chris's sis

@robjohn Oh, right! Thanks!

$$\sum_{n=1}^{\infty}\left(n-\sum_{k=2}^{n}\zeta{(k)}\right)=\frac{\pi^2}{6}$$

It's really cute now! :D

Jasper Loy

19:38

I answered 2 lhf just now.

robjohn

$$
\begin{align}
\sum_{n=1}^\infty\left(n-\sum_{k=2}^n\zeta(k)\right)
&=\sum_{n=1}^\infty\left(1-\sum_{k=2}^n(\zeta(k)-1)\right)\\
&=\sum_{n=1}^\infty\sum_{k=n+1}^\infty(\zeta(k)-1)\\
&=\sum_{k=2}^\infty\sum_{n=1}^{k-1}(\zeta(k)-1)\\
&=\sum_{k=2}^\infty(k-1)(\zeta(k)-1)\\
&=\sum_{k=2}^\infty\sum_{n=2}^\infty\frac{k-1}{n^k}\\
&=\sum_{n=2}^\infty\sum_{k=2}^\infty\frac{k-1}{n^k}\\
&=\sum_{n=2}^\infty\frac1{n^2}\sum_{k=0}^\infty\frac{k+1}{n^k}\\
&=\sum_{n=2}^\infty\frac1{n^2}\frac1{(1-1/n)^2}\\
&=\zeta(2)

Chris's sis

@robjohn nice :-)

Because ... $$\sum_{k=2}^\infty(\zeta(k)-1)=1$$

robjohn

Heh... my high scoring answer today is from Aug 20 last year. I wonder why the sudden interest.

@Chris'ssis yep

Jasper Loy

@robjohn Maybe there was a bump?

robjohn

@JasperLoy I don't see any edits.

Jasper Loy

19:49

@robjohn Haha, then maybe someone linked to it from a recent question...

robjohn

@JasperLoy I was looking into that...

Ian Mateus

@robjohn Mhenni linked his answer, a scrolling shows yours

robjohn

@IanMateus a current answer?

Ian Mateus

@robjohn see here

robjohn

@IanMateus I should have started at the bottom of this page

Ian Mateus

19:58

@robjohn Hehehe

robjohn

@IanMateus Now I should try to answer that question :-)

Ian Mateus

@robjohn definitely!

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